Infinite series - Inverse trigonometry

1. Aug 18, 2013

Saitama

1. The problem statement, all variables and given/known data
The sum of the infinite terms of the series
$$\text{arccot}\left(1^2+\frac{3}{4}\right)+\text{arccot}\left(2^2+\frac{3}{4}\right)+\text{arccot}\left(3^2+\frac{3}{4}\right)+............$$
is equal to
A)arctan(1)
B)arctan(2)
C)arctan(3)
D)arctan(4)

Ans: B

2. Relevant equations

3. The attempt at a solution
The given series can be written as
$$\sum_{r=1}^{\infty} \text{arccot}\left(r^2+\frac{3}{4}\right)$$
Simplifying the term inside the summation, I rewrite it as
$$\arctan\left(\frac{4}{4r^2+3}\right)$$
I don't see how to proceed further. I am thinking of converting the above in the form $\arctan\left(\frac{x-y}{1+xy}\right)$, rewrite it in the form $\arctan(x)-\arctan(y)$ and perform the summation but I can't see a way to do this.

Any help is appreciated. Thanks!

Last edited: Aug 18, 2013
2. Aug 18, 2013

vela

Staff Emeritus
I haven't worked it out, but I think you're on the right track. Try rewriting it as
$$r^2 + \frac{3}{4} = \left(r^2 - \frac{1}{4}\right) + 1 = \left(r+\frac{1}{2}\right)\left(r-\frac{1}{2}\right) + 1$$

3. Aug 18, 2013

Saitama

Wow. That's great. Thanks a lot vela!

I did try rewriting 3/4 as 1-1/4 but then I didn't notice that it could simplified to (r-1/2)(r+1/2).

4. Aug 19, 2013

verty

This is one of the most sadistic questions I've seen. Where are you getting these questions?

5. Aug 19, 2013

Saitama

From practice sheets. Do you want me to send the links to those practice sheets by PM? I don't know if its ok to post them here.

6. Aug 19, 2013

verty

A PM will be fine, thanks. I haven't seen these type questions before, is all.